Taking the usual definition of the propagator for a massless scalar field and taking the reciprocal:

$$f(x,y)\propto\left(\int \phi(x)\phi(y) e^{i\int \sqrt{-g}g^{\mu\nu}(z)\partial_\mu \phi(z)\partial_\nu\phi(z) d^4z}D\phi\right)^{-1}$$

When $g$ is the Minkowski metric we have that $f(x,y)\propto|x-y|^2$.

This suggests that the function $f(x,y)$ gives the proper time between two points traced by the shortest path. So if we let $g(x)$ become the metric of a general curved space-time, this should still hold true. We might write $f(x,y) = MaxTime(x,y)^2$

However.... in such cases as when $g$ is the field of a gravitional mass, there may be more than one locally minimum path between two space-time points, in which case this couldn't be true. Hence either the integral cannot be valid for this metric or the result is something diffent such as the average of all the proper times-squared for locally minimum paths.

Also for the Minkowski case we can say that when $f(x,y)>0$ the points are space-like separated and $f(x,y)<0$ the points are time-like separated. Does $f$ give us similar information when $g$ is a general curved space (perhaps with singularities at gravitional sources?)

Edit: To clarify, I meant $x$ and $y$ to be points in 3+1 dimensional space-time. Which I think was understood.

  • $\begingroup$ I may be making some idiotic mistake. But here goes anyway....So you are suggesting that field theoretic propagator $f(x,y)$ is related to the proper time measured along the minimum proper time path between $x$ and $y$, given the metric... So, basically, $f(x,y)$ is related to the path that a classical point particle would take between $x$ and $y$ given the metric? That does not seem right somehow. $\endgroup$ – insomniac Mar 20 at 10:39
  • $\begingroup$ @insomniac Yes, that's what I'm saying. A classical path is after all just the most likely quantum path. $\endgroup$ – zooby Mar 20 at 21:10
  • $\begingroup$ May be relevant : physics.stackexchange.com/questions/155076/… $\endgroup$ – insomniac Mar 21 at 9:02

My original answer was going to be much longer, but everything started to be a bit too long and complex, so here's some rough elements to give you how to derive the idea :

Formally, the Hadamard function (which is related to the Green's function by some differing contour integral) is defined on curved spacetimes by

$$G^{(1)}(x, y) = \sum_{\gamma} \frac{\Delta_\gamma^{\frac{1}{2}}(x,y)}{4\pi^2} \left[ \frac{1}{\sigma_{\gamma}(x,y)} + v_\gamma(x,y) \ln(|\sigma_\gamma(x,y)|) + \varpi_\gamma(x,y) \right]$$

Where we sum over every possible geodesic connecting two points. $\Delta$ is the van Vleck determinant, $\sigma$ the geodetic distance between $x$ and $y$, and $v$ and $\varpi$ are non-singular functions defined by the theory.

The Feynmann propagator is defined similarly, although the functions within that sum differ. So the solution is to take the sum of those geodesics.

We also have that the relation between the Feynman propagator and the relativistic point particle is

\begin{eqnarray} G_F(x, y) &=& \langle T[ \phi(x), \phi(y) ] \rangle\\ &=& \int \mathcal{D}\phi\ \phi(x) \phi(y) e^{i S[\phi]}\\ &=& \int_0^\infty \frac{dT}{T} \int_{x(0) = x}^{x(T) = y} \mathcal{D}x \exp\left[\int_0^T d\tau (g(\dot{x}, \dot{x}) + m^2) \right] \end{eqnarray}

There's a bunch of additional things to do for this to work properly, ie adding the einbein, ghost fields and so on, but this is roughly the relation we have between the two, and both evaluate to the appropriate propagator. The trick to do this, and I suspect, for the non-unique geodesic case, is to use the decomposition of the action into the extremal solution and variations, ie

\begin{eqnarray} x(\tau) = x_C(\tau) + y(\tau) \end{eqnarray}

So that $x_C(\tau)$ is the classical solution that extremize the action, and $y$ is any function on $[0,T]$ with $y(0) = y(T) = 0$. In the case of a cut locus, there are multiple such solutions.

Since we are dealing with massless fields here, for the flat space case we just have

\begin{eqnarray} G_F(x,y) = \frac{i}{8\pi^2 \sigma(x,y)} - \frac{1}{8\pi} \delta(\sigma(x, y)) \end{eqnarray}

or, for the Hadamard function,

\begin{eqnarray} G^{(1)}(x,y) = -\frac{1}{4\pi^2 \sigma(x,y)} \end{eqnarray}

Now let's consider the case of the cylinder. The Minkowski cylinder is the manifold $\mathbb{R} \times S$, with metric

\begin{equation} ds^2 = -dt^2 + d\theta^2 \end{equation}

On the cylinder, the propagator evaluates to

\begin{equation} G_C(x,y) = \sum_{k \in \mathbb{Z}} G(x, y + (0, kL)) \end{equation}

This changes the geodetic interval by

\begin{eqnarray} \sigma(x, y + (0, kL)) &=& |x - y - (0, kL)|^2\\ &=& -(x_t - y_t)^2 + (x_\theta - y_\theta - kL)^2 \end{eqnarray}

This is very much equivalent to the geodetic interval of the geodesic connecting $x$ to $y$ with winding number $k$, which corresponds indeed to every geodesics connecting the two. On the cylinder, two points are connected by the helix

\begin{eqnarray} \gamma(\lambda) &=& (a \lambda + t_0, b \lambda + \theta_0) \end{eqnarray}

at $\gamma(0) = x$, $t_0 = x_t$ and $\theta_0 = x_\theta$, and at $\gamma(1) = y$, $a + x_t = y_t$ and $b + x_\theta = y_\theta + k L$, so that the geodesic with winding number $k$ has the geodesic interval

\begin{eqnarray} \sigma_{\gamma_k}(x, y) &=& \int_0^1 -(x_t - y_t)^2 + (x_\theta - y_\theta - kL)^2 d\lambda\\ &=& -(x_t - y_t)^2 + (x_\theta - y_\theta - kL)^2 \end{eqnarray}

I attempted to do the case of the sphere, but I'm afraid this is getting a bit long for a short answer. This should be roughly equivalent, considering that on a sphere, the geodesics are parametrized by 1) the winding number 2) if the points are antipodal, the azimuthal angle. You can check (in Birrell & Davies, for instance) that in the case of the Einstein static universe (roughly a static sphere spacetime) that similarly to the cylinder, the propagator depends on an infinite sum over the winding number. I assume that in addition, there is an integral over $\varphi$ in the antipodal case that is already worked out here.

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  • $\begingroup$ So to summarise, it would be true to say the function $f(x,y)$ is a function of the proper times of all possible geodesic paths connecting two space-time points? Something like this: $1/(1/\tau_1 + 1/\tau_2+...)$ That would make sense, since the geodesic times would be the only invariants I can think of. $\endgroup$ – zooby Mar 20 at 21:28
  • $\begingroup$ Thanks for your detailed answer. It's good to know that sometimes my intuition is correct even though mostly its wrong! $\endgroup$ – zooby Mar 20 at 21:40
  • $\begingroup$ Just thinking a gravitational mass like a black hole would be similar to the cylindrical solution. The different geodesics being how many times the path orbits the black hole. $\endgroup$ – zooby Mar 20 at 21:55
  • $\begingroup$ It would be nice if there was some function $f(x,y,q)=\sum \tau_n q^n$ where $n$ was the winding number. To separate all the different maxima. $\endgroup$ – zooby Mar 21 at 0:26

Expanding on comment above : The reason the Flat spacetime propagator $f_{Mink}(x,y)$ has to go as some function of $ |x-y|$ can be boiled down to two facts :

$(1)$ The theory is translation invariant. This implies that $f_{Mink}$ can only be a function of $x-y$. No other dependence on $x$ and $y$ is allowed

$(2)$ The theory is Lorentz invariant, therefore, so must $f$. This implies that $f$ can only be a function of $|x-y|$.

However, your proposition effectively interprets $|x-y|$ as the proper time measured along the geodesic path between $x$ and $y$, which it is, but only accidentally in flat spacetime. There is no reason for there to be any relation between $f_g(x,y)$ and the proper time measured along the geodesic between $x$ and $y$ for a general metric $g$.

For one, such a relation would imply that as long as we have a non-interacting field theory in the massless limit (which is what you have) on a general curved spacetime, we may basically throw away the field theory itself , since the only relevant quantity it can produce (the propagator) can be found by solving the equations of motion for a classical point particle ? Why should we expect that to happen?

I feel that any relation between $f$ and the the geodesic is only possible when it is mandated by some symmetry requirement. This may be possible in other $g$'s apart from the Minkowski metric, but intuitively at least, I do not feel that this statement would hold generally.

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  • $\begingroup$ Well non-interacting field theories are not very interesting so why shouldn't most of it be "thrown away"? Actually finding the classical paths in curved space is non-trivial so it's not exactly making it any easier. $\endgroup$ – zooby Mar 20 at 21:22

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